1 Chapter 6 Gases 6.1 Properties of Gases 6.2 Gas Pressure Copyright © 2009 by Pearson Education, Inc.

2 Kinetic Theory of Gases A gas consists of small particles that move rapidly in straight lines. have essentially no attractive (or repulsive) forces. are very far apart. have very small volumes compared to the volume of the container they occupy. have kinetic energies that increase with an increase in temperature. Copyright © 2009 by Pearson Education, Inc.

3 Properties That Describe a Gas Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). Copyright © 2009 by Pearson Education, Inc.

4 Gas pressure is a force acting on a specific area. Pressure (P) = force area has units of atm, mmHg, torr, lb/in. 2, and kilopascals(kPa). 1 atm = 760 mm Hg (exact) 1 atm = 760 torr 1 atm = 14.7 lb/in. 2 1 atm = Pa 1 atm = kPa Gas Pressure

5 A. What is 475 mmHg expressed in atm? 1) 475 atm 2) atm 3) 3.61 x 10 5 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mmHg? 1) 2.00 mmHg 2) 1520 mmHg 3) mmHg Learning Check

6 Atmospheric Pressure Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth. Copyright © 2009 by Pearson Education, Inc.

7 Altitude and Atmospheric Pressure Atmospheric pressure is about 1 atmosphere at sea level. depends on the altitude and the weather. is lower at higher altitudes, where the density of air is less. is higher on a rainy day than on a sunny day. Copyright © 2009 by Pearson Education, Inc.

8 Barometer A barometer measures the pressure exerted by the gases in the atmosphere. indicates atmospheric pressure as the height in mm of the mercury column. Copyright © 2009 by Pearson Education, Inc.

9 A. The downward pressure of the Hg in a barometer is _____ than (as) the pressure of the atmosphere. 1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense than mercury. 2) H 2 O is heavier than mercury. 3) air is more dense than H 2 O. Learning Check

10 Chapter 6Gases 6.3 Pressure and Volume (Boyle’s Law) Copyright © 2009 by Pearson Education, Inc.

11 Boyle’s Law Boyle’s law states that the pressure of a gas is inversely related to its volume when T and n are constant. if volume decreases, the pressure increases. Copyright © 2009 by Pearson Education, Inc.

12 In Boyle’s law, the product P x V is constant as long as T and n do not change. P 1 V 1 = 8.0 atm x 2.0 L = 16 atm L P 2 V 2 = 4.0 atm x 4.0 L = 16 atm L P 3 V 3 = 2.0 atm x 8.0 L = 16 atm L Boyle’s law can be stated as P 1 V 1 = P 2 V 2 (T, n constant) PV Constant in Boyle’s Law

13 Solving for a Gas Law Factor The equation for Boyle’s law can be rearranged to solve for any factor. P 1 V 1 = P 2 V 2 Boyle’s law To solve for V 2, divide both sides by P 2. P 1 V 1 = P 2 V 2 P 2 P 2 V 1 xP 1 = V 2 P 2

14 Boyles’ Law and Breathing During an inhalation, the lungs expand. the pressure in the lungs decreases. air flows towards the lower pressure in the lungs. Copyright © 2009 by Pearson Education, Inc.

15 Boyles’ Law and Breathing During an exhalation, lung volume decreases. pressure within the lungs increases. air flows from the higher pressure in the lungs to the outside. Copyright © 2009 by Pearson Education, Inc.

16 Freon-12, CCl 2 F 2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mmHg after its pressure is changed to 2200 mmHg at constant T and n? 1. Set up a data table: Conditions 1Conditions 2 P 1 = 550 mmHgP 2 = 2200 mmHg V 1 = 8.0 LV 2 = Calculation with Boyle’s Law ?

17 2. When pressure increases, volume decreases. Solve Boyle’s law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 x P 1 P 2 V 2 = 8.0 L x 550 mmHg = 2.0 L 2200 mmHg pressure ratio decreases volume Calculation with Boyle’s Law (continued)

18 Learning Check For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant). 1) pressure decreases 2) pressure increases Copyright © 2009 by Pearson Education, Inc.

19 Learning Check If a sample of helium gas has a volume of 120 mL and a pressure of 850 mmHg, what is the new volume if the pressure is changed to 425 mmHg? 1) 60 mL 2) 120 mL3) 240 mL

20 If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 LB) 6.4 LC) 12.8 L Learning Check

21 Learning Check A sample of oxygen gas has a volume of 12.0 L at 600. mmHg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg Copyright © 2009 by Pearson Education, Inc.

22 Chapter 6Gases 6.4 Temperature and Volume (Charles’s Law) Copyright © 2009 by Pearson Education, Inc.

23 Charles’s Law In Charles’s Law, the Kelvin temperature of a gas is directly related to the volume. P and n are constant. when the temperature of a gas increases, its volume increases. Copyright © 2009 by Pearson Education, Inc.

24 Charles’s Law: V and T For two conditions, Charles’s law is written V 1 = V 2 (P and n constant) T 1 T 2 Rearranging Charles’s law to solve for V 2 : T 2 xV 1 = V 2 x T 1 T 1 V 2 = V 1 x T 2 T 1

25 Learning Check Solve Charles’s law expression for T 2. V 1 = V 2 T 1 T 2

26 A balloon has a volume of 785 mL at 21 °C. If the temperature drops to 0 °C, what is the new volume of the balloon (P constant)? 1.Set up data table: Conditions 1Conditions 2 V 1 = 785 mLV 2 = ? T 1 = 21 °C = 294 KT 2 = 0 °C = 273 K Be sure to use the Kelvin (K) temperature in gas calculations. Calculations Using Charles’s Law

27 Calculations Using Charles’s Law (continued) 2. Solve Charles’s law for V 2 : V 1 = V 2 T 1 T 2 V 2 = V 1 x T 2 T 1 V 2 = 785 mL x 273 K = 729 mL 294 K

28 A sample of oxygen gas has a volume of 420 mL at a temperature of 18 °C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443 °C 2) 170 °C 3) - 82 °C Learning Check

29 Use the gas laws to complete each sentence with 1) increases or 2) decreases. A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12 L to 24 L. D. Volume _______when T changes from 15 °C to 45 °C. Learning Check

30 Use the gas laws to complete each sentence with 1) increases or 2) decreases. A. Pressure 1) increases when V decreases. B. When T decreases, V 2) decreases. C. Pressure 2) decreases when V changes from 12 L to 24 L. D. Volume 1) increases when T changes from 15 °C to 45 °C. Solution

31 Chapter 6Gases 6.4 Temperature and Volume (Charles’s Law) Copyright © 2009 by Pearson Education, Inc.

32 Charles’s Law In Charles’s Law, the Kelvin temperature of a gas is directly related to the volume. P and n are constant. when the temperature of a gas increases, its volume increases. Copyright © 2009 by Pearson Education, Inc.

33 Charles’s Law: V and T For two conditions, Charles’s law is written V 1 = V 2 (P and n constant) T 1 T 2 Rearranging Charles’s law to solve for V 2 : T 2 xV 1 = V 2 x T 1 T 1 V 2 = V 1 x T 2 T 1

34 Learning Check Solve Charles’s law expression for T 2. V 1 = V 2 T 1 T 2

35 A balloon has a volume of 785 mL at 21 °C. If the temperature drops to 0 °C, what is the new volume of the balloon (P constant)? 1.Set up data table: Conditions 1Conditions 2 V 1 = 785 mLV 2 = ? T 1 = 21 °C = 294 KT 2 = 0 °C = 273 K Be sure to use the Kelvin (K) temperature in gas calculations. Calculations Using Charles’s Law

36 Calculations Using Charles’s Law (continued) 2. Solve Charles’s law for V 2 : V 1 = V 2 T 1 T 2 V 2 = V 1 x T 2 T 1 V 2 = 785 mL x 273 K = 729 mL 294 K

37 A sample of oxygen gas has a volume of 420 mL at a temperature of 18 °C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443 °C 2) 170 °C 3) - 82 °C Learning Check

38 Use the gas laws to complete each sentence with 1) increases or 2) decreases. A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12 L to 24 L. D. Volume _______when T changes from 15 °C to 45 °C. Learning Check

39 Use the gas laws to complete each sentence with 1) increases or 2) decreases. A. Pressure 1) increases when V decreases. B. When T decreases, V 2) decreases. C. Pressure 2) decreases when V changes from 12 L to 24 L. D. Volume 1) increases when T changes from 15 °C to 45 °C. Solution

40 Chapter 6Gases 6.5 Temperature and Pressure (Gay-Lussac’s Law) Copyright © 2009 by Pearson Education, Inc.

41 Gay-Lussac’s Law: P and T In Gay-Lussac’s law the pressure exerted by a gas is directly related to the Kelvin temperature. V and n are constant. P 1 = P 2 T 1 T 2 Copyright © 2009 by Pearson Education, Inc.

42 Learning Check Solve Gay-Lussac’s law for P 2. P 1 = P 2 T 1 T 2

43 A gas has a pressure at 2.0 atm at 18 °C. What is the new pressure when the temperature is 62 °C? (V and n constant) 1. Set up a data table: Conditions 1Conditions 2 P 1 = 2.0 atmP 2 = T 1 = 18 °C T 2 = 62 °C = 291 K = 335 K Calculation with Gay-Lussac’s Law ?

44 Calculation with Gay-Lussac’s Law (continued) 2. Solve Gay-Lussac’s law for P 2 : P 1 = P 2 T 1 T 2 P 2 = P 1 x T 2 T 1 P 2 = 2.0 atm x 335 K = 2.3 atm 291 K temperature ratio increases pressure

45 Learning Check A gas has a pressure of 645 torr at 128 °C. What is the temperature in Celsius if the pressure increases to 824 torr? (n and V remain constant)

46 Solution 2. Solve Gay-Lussac’s law for T 2 : P 1 = P 2 T 1 T 2 T 2 = T 1 x P 2 P 1 T 2 = 401 K x 1140 torr = 709 K = 436 °C 645 torr pressure ratio increases temperature

47 Chapter 6 Gases 6.6 The Combined Gas Law Copyright © 2009 by Pearson Education, Inc.

48 The combined gas law uses Boyle’s law, Charles’s law, and Gay-Lussac’s law (n is constant). P 1 V 1 =P 2 V 2 T 1 T 2 Combined Gas Law

49 A gas has a volume of 675 mL at 35 °C and atm pressure. What is the volume (mL) of the gas at -95 °C and a pressure of 802 mmHg? (n constant) Learning Check

50 A sample of helium gas has a volume of L, a pressure of atm, and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm? (n is constant) 1. Set up data table. Conditions 1Conditions 2 P 1 = atm P 2 = 3.20 atm V 1 = L (180 mL) V 2 = 90.0 mL T 1 = 29 °C = 302 KT 2 = ?? Combined Gas Law Calculation

51 Chapter 6 Gases 6.7 Volume and Moles (Avogadro’s Law) Copyright © 2009 by Pearson Education, Inc.

52 Avogadro's Law: Volume and Moles In Avogadro’s law the volume of a gas is directly related to the number of moles (n) of gas. T and P are constant. V 1 = V 2 n 1 n 2 Copyright © 2009 by Pearson Education, Inc.

53 Learning Check If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure? 1) 0.94 L 2)1.8 L 3) 2.4 L Copyright © 2009 by Pearson Education, Inc.

54 The volumes of gases can be compared at STP, Standard Temperature and Pressure, when they have the same temperature. standard temperature (T) 0 °C or 273 K the same pressure. standard pressure (P) 1 atm (760 mmHg) STP

55 Molar Volume At standard temperature and pressure (STP), 1 mole of a gas occupies a volume of 22.4 L, which is called its molar volume. Copyright © 2009 by Pearson Education, Inc.

56 Molar Volume as a Conversion Factor The molar volume at STP can be used to write conversion factors L and 1 mole 1 mole 22.4 L

57 Using Molar Volume What is the volume occupied by 2.75 moles of N 2 gas at STP? The molar volume is used to convert moles to liters moles N 2 x 22.4 L = 61.6 L 1 mole Copyright © 2009 by Pearson Education, Inc.

58 A. What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many g of He are present in 8.00 L of gas at STP? 1) 25.6 g2) g3) 1.43 g Learning Check

59 Gases in Equations The volume or amount of a gas at STP in a chemical reaction can be calculated from STP conditions. mole factors from the balanced equation.

60 STP and Gas Equations What volume (L) of O 2 gas at STP is needed to completely react with 15.0 g of aluminum? 4Al(s) + 3O 2 (g) 2Al 2 O 3 (s) Plan: g Al mole Al mole O 2 L O 2 (STP) 15.0 g Al x 1 mole Al x 3 moles O 2 x 22.4 L (STP) 27.0 g Al 4 moles Al 1 mole O 2 = 9.33 L of O 2 at STP

61 What mass of Fe will react with 5.50 L of O 2 at STP? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Learning Check

62 Chapter 6 Gases 6.8 Partial Pressures (Dalton’s Law) Copyright © 2009 by Pearson Education, Inc.

63 The partial pressure of a gas is the pressure of each gas in a mixture. is the pressure that gas would exert if it were by itself in the container. Partial Pressure

64 Dalton’s law of partial pressures indicates that pressure depends on the total number of gas particles, not on the types of particles. the total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases. P T = P 1 + P 2 + P Dalton’s Law of Partial Pressures

65 Dalton’s Law of Partial Pressures Copyright © 2009 by Pearson Education, Inc.

66 For example, at STP, 1 mole of a pure gas in a volume of 22.4 L will exert the same pressure as 1 mole of a gas mixture in 22.4 L. V = 22.4 L Gas mixtures Total Pressure 0.5 mole O mole He 0.2 mole Ar 1.0 mole 1.0 mole N mole O mole He 1.0 mole 1.0 atm Copyright © 2009 by Pearson Education, Inc.

67 Scuba Diving When a scuba diver dives, the increased pressure causes N 2 (g) to dissolve in the blood. If a diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends." Helium, which does not dissolve in the blood, is mixed with O 2 to prepare breathing mixtures for deep descents. Copyright © 2009 by Pearson Education, Inc.

68 Learning Check A scuba tank contains O 2 with a pressure of atm and He at 855 mmHg. What is the total pressure in mmHg in the tank? Copyright © 2009 by Pearson Education, Inc.

69 For a deep dive, a scuba diver uses a mixture of helium and oxygen with a pressure of 8.00 atm. If the oxygen has a partial pressure of 1280 mmHg, what is the partial pressure of the helium? 1) 520 mmHg 2) 2040 mmHg 3) 4800 mmHg Learning Check

70 Gases We Breathe The air we breathe is a gas mixture. contains mostly N 2 and O 2, and small amounts of other gases. Copyright © 2009 by Pearson Education, Inc.

71 A. If the atmospheric pressure today is 745 mmHg, what is the partial pressure (mmHg) of O 2 in the air? 1) ) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mmHg) N 2 in the air? 1) 557 2) 9.143) Learning Check

72 Blood Gases In the lungs, O 2 enters the blood, while CO 2 from the blood is released. In the tissues, O 2 enters the cells, which releases CO 2 into the blood. Copyright © 2009 by Pearson Education, Inc.

73 Blood Gases In the body, O 2 flows into the tissues because the partial pressure of O 2 is higher in blood, and lower in the tissues. CO 2 flows out of the tissues because the partial pressure of CO 2 is higher in the tissues, and lower in the blood. Partial Pressures in Blood and Tissue Oxygenated Deoxygenated Gas BloodBlood Tissues O or less CO or greater

74 Gas Exchange During Breathing Copyright © 2009 by Pearson Education, Inc.